Let g be a group of order 42. Find the subgroup of order 6 with sylow

To find a subgroup of order 6 in a group \( G \) of order 42 using Sylow's theorems, we proceed as follows:

1. **Determine the prime factorization of the group order:**
   \[
   |G| = 42 = 2 \times 3 \times 7
   \]

2. **Identify the Sylow \( p \)-subgroups for each prime \( p \):**
   - **Sylow 2-subgroups**: A Sylow 2-subgroup has order \( 2^1 = 2 \). Let \( n_2 \) denote the number of Sylow 2-subgroups. According to Sylow's theorems, \( n_2 \) divides \( 42/2 = 21 \) and \( n_2 \equiv 1 \pmod{2} \). Thus, \( n_2 \) can be 1 or 21.
   - **Sylow 3-subgroups**: A Sylow 3-subgroup has order \( 3^1 = 3 \). Let \( n_3 \) denote the number of Sylow 3-subgroups. According to Sylow's theorems, \( n_3 \) divides \( 42/3 = 14 \) and \( n_3 \equiv 1 \pmod{3} \). Thus, \( n_3 \) can be 1 or 7.
   - **Sylow 7-subgroups**: A Sylow 7-subgroup has order \( 7^1 = 7 \). Let \( n_7 \) denote the number of Sylow 7-subgroups. According to Sylow's theorems, \( n_7 \) divides \( 42/7 = 6 \) and \( n_7 \equiv 1 \pmod{7} \). Thus, \( n_7 \) can only be 1.

Since \( n_7 = 1 \), there is a unique Sylow 7-subgroup in \( G \). Let's denote this subgroup by \( P_7 \). Because \( P_7 \) is unique, it is normal in \( G \).

3. **Find the intersection and product of Sylow subgroups:**
   To form a subgroup of order 6, we need to combine a Sylow 2-subgroup and a Sylow 3-subgroup. Let's denote the Sylow 2-subgroup by \( P_2 \) and the Sylow 3-subgroup by \( P_3 \).

4. **Combine Sylow 2-subgroup and Sylow 3-subgroup:**
   If \( P_2 \) and \( P_3 \) intersect trivially, their product will have the desired order. The order of the product \( P_2P_3 \) is given by:
   \[
   |P_2P_3| = \frac{|P_2| \cdot |P_3|}{|P_2 \cap P_3|}
   \]
   Since \( P_2 \) and \( P_3 \) intersect trivially (i.e., \( |P_2 \cap P_3| = 1 \)), we get:
   \[
   |P_2P_3| = |P_2| \cdot |P_3| = 2 \times 3 = 6
   \]

Thus, the product of a Sylow 2-subgroup and a Sylow 3-subgroup forms a subgroup of order 6 in \( G \).

### Summary:
By applying Sylow's theorems, we conclude that \( G \) contains subgroups of order 6, which are the products of its Sylow 2-subgroups and Sylow 3-subgroups. Specifically, these subgroups can be constructed as the product of any Sylow 2-subgroup and any Sylow 3-subgroup, assuming they intersect trivially.

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. How many Sylow 2-subgroups can exist in a group of order 42?
2. What is the significance of a normal Sylow subgroup?
3. Can a Sylow \( p \)-subgroup be non-normal?
4. How can we determine the number of Sylow 3-subgroups if \( n_3 = 7 \)?
5. What are the possible structures of a group of order 42?
6. How do the Sylow theorems help in understanding the composition of a group?
7. What are the general properties of Sylow subgroups in finite groups?
8. Can there be more than one subgroup of a given order in a group of order 42?

**Tip:** Always check for normality of subgroups when applying Sylow theorems, as normal subgroups often simplify the structure analysis of a group.

Learn how to find a subgroup of order 6 in a group of order 42 using Sylow's theorems. Understand the prime factorization of group order, identify Sylow subgroups for primes 2, 3, and 7, and combine Sylow 2-subgroups and Sylow 3-subgroups to form the desired subgroup. Suitable for advanced undergraduate students studying group theory and algebraic structures.

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